34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 1, 21 bp ≙ 8 rp; f = 0.38095
1.1.
Coordination number in the hexagonal face centred cell: 12
Coordination number in the cubic body centred cell: 8
1.2. Lattice constant:
334.4 pm
Berechnung
4r  a 3
4  144.8
a
 334.4 pm
3
1.3. Molar mass, element:
Calculation
2 bp
2 bp
47.9 g/mol , titanum
4 bp
VZelle  ( 334.4  10 10 )3  3.74  10 23 cm3

2 M( X )
  N A  VZelle 4.506  0.944  6.022  10 23  3.74  10 23
M( X ) 

 47.9 g / mol
N A  VZelle
2
2
1.4.
Compound A:
FeTiO3
2bp
Compound B:
TiOSO4
2bp
Compound C:
TiO2
1bp
Compound E:
TiCl3
1bp
Compound F:
[TiCl4(P(C2H5)3)2] 1bp
Compound G:
Ti(O2)SO4
1bp
1.5. Geometry:
Compound D:
TiCl4
1bp
octahedral
1.6. Reaction equation:
1 bp
2 Ti2+ + 2 H2O → 2 Ti3+ + H2 + 2 OH-
1.7. Reasons for corrosion resistance: passivation by TiO2
1
2 bp
1 bp
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 2, 19 bp ≙ 7 rp; f = 0.36842
2.1. Reaction equation:
2.2.
Fe2O3 + 6 HCl → 2 FeCl3 + 3 H2O
Reducing agents
Zn, Sn2+, SO321,5 bp
2.3. Reaction equation:
1 bp
Reasoning
E° (RM) < 0.77 V
6 Fe2+ + Cr2O72- + 14 H+ → 6 Fe3+ + 2 Cr3+ + 7 H2O
1 bp
1 bp
2.4. Mass fraction of Fe2O3: n( Fe2  )  6  n( Cr O 2  )  6  0.1  33.74  10 3 mol
2 7
3 bp
m( Fe2O3 )  0.5  n( Fe2  )  159.7  1.616 g
w
1,616
 100  40.4%
4
2.5. Equilibrium constant K: E   1.33  0.77  0.56V
K e
6.F .E 
R .T
e
6 964850.56
8.314298
2.6. Potential at the equivalence point:
Eeq  
2.7.
Redox indicators
iron phenanthroline, ferroine
2 bp
 6.7  10 56
6  1.33  0.77
 1.25V
7
1bp
1.5 bp
Reasoning
E° (Ind) is in the region of Eeq 1 bp
2.8. no, because E°(Cl2/Cl-) > E°(Cr2O72-/Cr3+)
1 bp
2.9. minimum pH-value, at which Cl2 will be produced:
 
0.059
lg H 
5
(
1
.
36

1.51 )  5
lg H  
0.059  8

1.59
H  10
1.36  1.51 
8
 
 
pH  1.59
3 bp
2.10.
Change of standard potentials: they decrease
1 bp
2.11.
Effect of the addition of phosphoric acid: complexation of Fe3+
1 bp
2
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 3, 19 bp ≙ 7 rp; f = 0.36842
3.1. Balanced reaction equations:
3 bp
C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O (l)
C2H6O(l) + + 3 O2(g) → 2 CO2(g) + 3 H2O (l)
C12H22O11(s) → C12H22O11(aq)
C2H6O(l) → C2H6O(aq)
C12H22O11(aq) + H2O (l) → 4 C2H6O(aq) + 2 CO2(g)
3.2. ΔRH⊖ = -218 kJ
(1)
(2)
(3)
(4)
(5)
3 bp
Calculation
(5) = (1) – (3) – 4(B) + 4(D) = -5645 – 5 + 4‧1368 - 4‧10 = - 218
3.3. Balanced reaction equation: 2 NO2 → 2 NO + O2
3.4. order of reaction: 2nd order
1 bp
k = 0.54 L/mol.s
Calculation:
v = k‧[NO2]x
z.B.:
k1 =
k2 =
k3 =
k4 =
⇒ log v = log k + x‧log [NO2]
log 5.4‧10-5 = log k + x‧log 0.010 und log 1.38‧10-4 = log k + x‧log 0.016 ⇒
0.4075 = x‧0.0204
x≈2
k = v/[NO2]2
5.4‧10-5/0.012 = 5.40‧10-1
7.78‧10-5/0.0122 = 5.35‧10-1
1.57‧10-4/0.0142 = 5.41‧10-1
2.05‧10-4/0.0162 = 5.39‧10-1
3 bp
3.5. ∆RH⊝ = 114.2 kJ
1 bp
∆RS⊝ = 145 J/K
1 bp
Berechnung:
∆RH⊝ = 2‧90.3 - 2‧33.2 = 114.2 kJ
∆RS⊝ = 2‧211 + 205 - 2‧241 = 145 J/K
∆RG⊝ = 114.2 - 283‧0.145 = 73.2 kJ
3
∆RG⊝ = 73,2 kJ
1 bp
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
3.6. T ≧ 788 K
1.5 bp
Calculation
∆RG⊝ = 0 implies that the equilibrium is located in the middle (on the right side before)
T = ∆RH⊝/∆RS⊝ = 787.6
3.7. a: to the right
0,5 bp
b: to the left
0,5 bp
c: to the left
3.8. p(O2) ≦ 0.44 bar
3 bp
Calculation:
 RG   RG   RT  ln
p ( NO ) p ( O2 )
p ( NO2 )2
0
RT  ln 0.01 1 p2 ( O2 )  114200  500  145   41700
2
ln p( O2 )  
0.5 bp
41700
 ln( 0.01 )2  0.8209 
8.314  500
4
p( O2 )  e 0.8209  0.44
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 4, 22 bp ≙ 8 rp; f = 0.36364
4.1. Equation:
O2NC6H4OCOCH3 + H2O → O2NC6H4OH + CH3COOH
4.2. ε = 14560 L/mol.cm
1 bp
2 bp
Calculation:
All of p-nitrophenyl acetate has been converted: ε =1.456/10-4
4.3. Proof:
At = A(PNP) + A(PNA) = ε(PNP)c(PNP) + ε(PNA)c(PNA) = [ε(PNP)(c0- c(PNA)) + ε(PNA)c(PNA)]
At = A∞ + c(PNA)[ ε(PNA) - ε(PNP)] or
At = A∞ - c(PNA)[ -ε(PNA) + ε(PNP)]
A∞- At = [ε(PNP) - ε(PNA)]‧ c(PNA)
5 bp
4.4. Order : 1st order
k = 3.19‧10-4 s-1
Table :
t (s)
300
900
1500
3000
4500
6000
A∞ - At
1.304
1.079
0.903
0.570
0.356
0.212
ln(A∞ - At)
0.265
0.076
-0.102
-0.562
-1.033
-1.551
1/(A∞ - At)
0.767
0.927
1.107
1.754
2.809
4.717
Strategy to find out graphically:
0 order: A∞ - At versus t should be a straight line, slope: -k
1st order: ln(A∞ - At) versus t should be a straight line, slope: -k
2nd order: 1/(A∞ - A) versus t should be a straight line, slope: k
with graphics quality required: 8 bp
5
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
⋆
(⋆)ln(E∞-Et)
(⋆)0.2
↑
⋄
⋆
(⋆)0.0
(⋄)4.0
⋆
∙
(∙)1.3
(∙)E∞-Et ↑
(⋄)3.0
⋄
(⋆)-0.5
∙
⋆
(∙)1.0
∙
(⋄)2.0
straight line
deviation of points
⋄
(⋆)-1.0
∙
⋆
(∙)0.5
⋄
(⋄)1.0
⋄
⋄
∙
deviation of points
∙
(⋆)-1,5
t (s) →
(⋄)1/(E∞-Et)
1000
↑
2000
3000
6
4000
5000
⋆
6000
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
4.5. Correspondence: equation of hydrolysis – order of reaction:
1 bp
The reaction behaves according to 1st-order because water is present in a huge surplus and the
concentration remains constant.
4.6. k = 3.9‧10-4 s-1
3 bp
Calculation:
k
1
( A  At )t 1
ln 
t2  t1 ( A  At )t 2
k1 = 3.89‧10-4
k1 = 3.94‧10-4 etc.
4.7. EA = 29.7 kJ
2 bp
Calculation:
k ( T2 ) E A  1
1
   
ln

k ( T1 )
R  T1 T2 
1
1
1
k ( T2 )
E A  R      ln
k ( T1 )
 T1 T2 
1
1 
3.9
 1
E A  8.314  

 29.7
  ln
3.2
 298 303 
7
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 5, 18 bp ≙ 7 rp; f = 0.38889
5.1. Tick the right box:
⊗
The concentration of iron is too low for an optimal determination, because the absorbance
expected will be too low.
1 bp
5.2. Concentration of the complex: 8.89•10-5 mol/L
2 bp
Calculation: Because of iron being present in surplus and bipyridyl is the limiting reagent, the
approximated concentration of the complex will be:
[[Fe(bipyr)3]2+]
5.3.
( 10
4
2.667  10 4
=
 8.89  10  5
3
x
 1017.58
4
3
 x )  ( 2.667  10  3 x )
3 bp
5.4. Concentration after the first iteration step: c = 8.798‧10-5 mol/L
Concentration after the second iteration step: c = 8.800‧10-5 mol/L
10 bp
8.89  10 5
 1017.58
( 10  4  8.89  10  5 )  ( 2.667  10  4  3 x )3
8.89  10 5
 ( 2.667  10  4  3 x )3
4
5
17.58
( 10  8.89  10 )  10
3
8.89  10 5
 2.667  10  4  3 x
( 10  4  8.89  10  5 )  1017.58
x=8.798‧10-5 mol/L
8.798  10 5
 1017.58
4
5
4
3
( 10  8.798  10 )  ( 2.667  10  3 x )
x= 8.801‧10-5 mol/L
5.5. A = 0.7372
Calculation:
2 bp
A = ε‧c‧d = 8.377‧103‧8.800‧10-5‧1 = 0.7372
8
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 6, 21 bp ≙ 7 rp; f = 0.33333
6.1. Reaction equation:
5 U4+ + 2 MnO4- + 2 H2O ⇌ 5 UO22+ + 2 Mn2+ + 4 H+
3 bp
6.2. Total concentration c = 0.0667 mol/L
3 bp
5  cKMnO4  VKMnO4  2  cU 4  VU 4
5‧0.02‧0.01334 = 2‧x‧0.01
x  cU 4  cPO3  0.0667 mol / L = ctotal P
4
6.3. Calculation:
[H3O+] = 10-6.5
[H3PO4] ≈ 0
[PO43-] ≈ 0
0.0667 = [H2PO4-] + [HPO42-]
7 bp
[ HPO42  ]  [ H 3O  ]
 10 7.21

[ H 2 PO4 ]
[ HPO42  ]  10 6.5
 10 7.21
2
0.0667  [ HPO4 ]
[HPO42-] = 0.0109 mol/L
[H2PO4-] = 0.0558 mol/L
[ H 2 PO4 ]  [ H 3O  ] 0.0558  10 6.5

 3.00  10 6
10  2.23
10  2.23
10 12.32  [ HPO42  ] 10 12.32  0 ,0109
[ PO43  ] 

 1.65  10  8
[ H 3O  ]
10 6.5
[ H 3 PO4 ] 
Neglect was correct!
6.4. Total concentration of calcium c = 3.32‧10-4 mol/L
1.00-26 = x3‧(1.65‧10-8)2
x3
1.00  10 26
 3.32  10  4 mol / L
8 2
( 1.65  10 )
9
2 bp
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
6.5. Calculation:
6 bp
cNaOH = cNaOH
n 2
  40  0.05 mol / L
V 1
assumption: reaction only with H2PO4- to give HPO42- →
[H2PO4-] = 0.05582 – 0.05 = 0.00582 mol/L
[HPO42-] = 0.0109 + 0.05 = 0.0609 mol/L
pH  pK A  lg
cS
0.00582
 7.21  lg
 8.23
cB
0.0609
[H3O+] = 10-8.23 = 5.89‧10-9 mol/L
[ H 3 PO4 ] 
[ PO43  ] 
[ H 2 PO4 ]  [ H 3O  ] 0 ,00582  5.89  10 9

 5.82  10  9
 2.23
 2.23
10
10
10 12.32  [ HPO42  ] 10 12.32  0.0609

 4.95  10 6
[ H 3O  ]
5.89  10  9
10
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 7, 25 bp ≙ 8 rp; f = 0.32000
7.1. Strucutres A to H
A: 1 bp
B: 1.5 bp
C: 2 bp
D: 2 bp
OH
O
OH
O
OH
OH
O
O
E: 1 bp
F: 1.5 bp
OH
G: 1bp
H: 1bp
Br
7.2. stereo descriptors
geraniol: E-
1 bp
nerol: Z-
7.3. structur of K+
1 bp
1.5 bp
OH2+
7.3. structur of L+
1.5 bp
+
7.3. structur of M+
2 bp
+
7.3. formulae + H2O
-H+
1 bp
11
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
7.4. centre
1 bp
7.5. N
1 bp
7.5. O
1 bp
OH
*
OH
O
OH
7.6. mechanism
3 bp
+
+ H2O
- H+
+
12
OH
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
Problem 8, 22 bp ≙ 8 rp; f = 0.36364
8.1. D-sorbit 1 bp
8.1. sorbose 2 bp
CH2OH
CH2OH
H
O
OH
HO
HO
H
H
OH
H
H
OH
HO
1 bp
OH
OH
4 bp
O
CH2OH
CH2OH OH
O
OH
OH
OH
OH
CH2OH
OH
H
8.3. stereochemical relationship 1 bp
enantiomeric
8.4. Haworth-formulae of D-sorbose
O
OH
CH2OH
CH2OH
8.2. membership:
L-series
H
CH2OH CH2OH
O
OH
CH2OH
OH
OH
OH
OH
8.5. IUPAC-name: (3R,4S,5R)-1,3,4,5,6-Pentahydroxyhexan-2-on
8.6. acetonide
3 bp
2 bp
O
O
O
HO
O
O
2 bp
8.7. lactone X
O
OH
OH
CH2OH
HO
O
O
H
CH2OH
13
34th Austrian Chemistry Olympiad
National Competition
Theoretical part – solution
June 14th, 2008
8.8. relation X-vitamin C
keto-enol-tautomerism
8.9. centres of chirality
1 bp
HO
number of stereoisomeres
O
HO
1 bp
O
1 bp
4 stereoisomeres
*
HO
H
*
CH2OH
8.10. acidic groups
HO
HO
HO
O
1 bp
O
H
CH2OH
8.11. conjugate base
1 bp
HO
-O
HO
8.12. oxidized ascorbic acid
O
O
O
O
HO
H
CH2OH
O
O
H
CH2OH
14
1 bp